Satellite internet · Session 3 of 8
Why the relay stays up
What speed does an orbit require, and where does that number come from?
Throw a ball level, as hard as you can. It lands. Throw it harder and it lands further away. Earth's surface, meanwhile, curves away beneath it: travel 7.9 km horizontally and the ground has dropped about 4.9 m — which is exactly how far anything falls in the first second. So there is a throwing speed, 7.9 km every second, at which the ground curves away as fast as the ball drops toward it. The ball keeps falling and never lands. That is not an analogy for an orbit. It is what an orbit is.
By the end
- Derive v = √(GM / r) by setting gravitational attraction equal to the centripetal requirement of circular motion.
- Evaluate orbital speed and period at 550 km and at geostationary altitude, and check the second against the length of a day.
- Build one consistent scale model and explain why altitude and orbital radius are not interchangeable.
The idea
Newton's law of universal gravitation gives the attraction between two masses as F = GMm / r², where r is measured from Earth's centre, not its surface, and GM for Earth is 3.986 × 10⁵ km³/s². The mass of the satellite appears on both sides of what follows and cancels, which is why a 300 kg spacecraft and a dropped spanner orbit identically.
Anything moving in a circle of radius r at speed v is accelerating toward the centre at v² / r — this is the centripetal acceleration, and it is a requirement, not a force. Something must supply it. In orbit that something is gravity, so GMm / r² = mv² / r, and after cancelling m and one factor of r you get v = √(GM / r). The whole of orbital mechanics starts here.
Put numbers in. At Starlink's main shell, 550 km up, r = 6371 + 550 = 6921 km, so v = √(398,600 / 6921) = 7.59 km/s — roughly 27,300 km/h. The orbital period is T = 2πr / v = 5,730 s = 95.5 minutes, so the satellite laps the planet about 15 times a day. Geostationary satellites sit at r = 42,164 km, where v drops to 3.07 km/s and T stretches to 23.93 hours — one sidereal day, which is precisely why they appear to hang still.
Notice the direction of the trade. Higher orbits are slower, not faster, because v falls as 1/√r. Squaring T = 2πr / √(GM / r) gives T² = 4π²r³ / GM, which is Kepler's third law: the square of the period goes as the cube of the radius. The same relation that places the planets places the shells of a satellite constellation.
Gravity has not gone anywhere up there. At 550 km, g = 9.81 × (6371 / 6921)² = 8.31 m/s², about 85% of its value in your kitchen. Astronauts float not because gravity is absent but because they and their spacecraft are falling together, so nothing presses on anything — free fall, which looks identical to weightlessness from inside.
Low orbits are not free. There is still enough atmosphere at 550 km to produce measurable drag, so satellites there carry thrusters and spend propellant maintaining altitude. When the propellant runs out, or the spacecraft fails, drag wins: an object at 550 km re-enters within a few years, while the same object at 1,200 km would stay up for centuries. That difference is the main reason low shells are now preferred.
Velocity, gravity and the curve they produce
Gather
Ruler, string, protractor, paper, pencil, calculator and a soft ball. Use a 25 cm paper Earth, or a 1 m outline if you have the floor space.
Build and investigate
- Draw an Earth circle and a circular orbit around it. At four positions, draw the satellite's instantaneous velocity along the tangent and its gravitational acceleration toward Earth's centre. These arrows are perpendicular in a circular orbit — check each one with a protractor.
- Mark with a dashed arrow where the satellite would go if gravity vanished. It is the tangent line, by Newton's first law. Say in one sentence what gravity is actually doing to the velocity: changing its direction, not its magnitude.
- Drop a soft ball, then throw it level into clear space. Both reach the floor in about the same time. Measure that time and the horizontal distance of the thrown one, then explain what would have to change for it not to land.
- Derive v = √(GM / r) from GMm / r² = mv² / r, showing the cancellation of m and of one r. Evaluate it at r = 6921 km, then find T = 2πr / v in minutes.
- Build the scale model with Earth's diameter taken as 12,742 km. On a 25 cm Earth, 550 km of altitude is 25 × 550 ÷ 12,742 = 1.08 cm. Mark it. Then mark geostationary altitude at the same scale and compare the two heights with a ruler.
- Label on your model which distances are measured from the surface (altitude) and which from the centre (orbital radius). Confirm that the r you used in the formula was the one from the centre.
Run the numbers
One substitution carries this entire lesson: gravity supplies the centripetal acceleration, and nothing else is needed.
| Relation | Expression | What it means |
|---|---|---|
| Gravitation | F = GMm / r² | GM = 3.986 × 10⁵ km³/s² for Earth |
| Circular motion | a = v² / r | acceleration required to curve, directed inward |
| Orbital speed | v = √(GM / r) | set the two equal; the satellite's mass cancels |
| Orbital period | T = 2πr / v | equivalently T² = 4π²r³ / GM (Kepler III) |
| Gravity with height | g(r) = g₀ · (R / r)² | R = 6371 km, g₀ = 9.81 m/s² |
| Radius and altitude | r = R + h | formulas take r; maps quote h |
Core · Grade 7–8
- Evaluate v = √(GM / r) at h = 550 km and convert to km/h. Then find T in minutes and the number of orbits per day. Compare your period with the 95.5 minutes quoted above.
- Calculate g at 550 km using g(r) = 9.81 × (6371 / 6921)². State the percentage of surface gravity, then write one sentence explaining why astronauts still float.
- Repeat the speed and period calculation for the ISS at 420 km. Is it faster or slower than the Starlink shell, and does that match v ∝ 1/√r?
Deeper · Grade 9–12
- Derive T² = 4π²r³ / GM from v = √(GM / r) and T = 2πr / v, showing every step. Then solve it for the r that makes T one sidereal day (86,164 s) and confirm you recover geostationary orbit at 42,164 km — an altitude of 35,786 km.
- Show that the ball in step 3 and a satellite obey the same equation. In one second a projectile falls ½ g t² = 4.9 m. Using the small-angle approximation, Earth's surface drops d² / 2R below the tangent over a horizontal distance d. Set the two equal and solve for d. Compare your answer with the orbital speed you calculated, and account for the difference.
- A satellite at 550 km and one at 1,200 km both lose power. Using the fact that atmospheric density falls roughly exponentially with altitude, explain why one re-enters in years and the other in centuries, and why a constellation operator might accept more drag deliberately.
Measure and record
For every distance, record its name, its real value in km, whether it is measured from the centre or the surface, the scale calculation and the model value with units. For every speed or period, record the formula, the substitution and the result with units.
Worked example
At 550 km: r = 6371 + 550 = 6921 km, so v = √(398,600 ÷ 6921) = √57.59 = 7.59 km/s, and T = 2π × 6921 ÷ 7.59 = 5,730 s = 95.5 min, giving 1440 ÷ 95.5 ≈ 15 orbits a day. Check the method against a known case before trusting it: at r = 42,164 km the same two formulas give v = 3.07 km/s and T = 86,164 s = 23.93 h, one sidereal day — which is the defining property of geostationary orbit, so the method is sound. On the scale model, 25 × 550 ÷ 12,742 = 1.08 cm and 25 × 35,786 ÷ 12,742 = 70.2 cm. A 25 cm Earth with the Starlink shell about a centimetre off its surface is the honest picture: these satellites are not far away, they are extremely close and moving extremely fast.
Pause and explain
A satellite in a circular 550 km orbit shuts its thrusters off completely. Predict what happens over the next hour, and over the next decade, and justify each with a mechanism.
Check your reasoning after your attempt
Over the next hour: essentially nothing changes. No thrust is needed to stay in orbit — gravity alone supplies the centripetal acceleration, and the satellite coasts through roughly two-thirds of a lap at 7.59 km/s. Over a decade: it re-enters. At 550 km there is still enough atmosphere to exert drag, which slowly removes energy; the orbit sinks, the air thickens, and the loss accelerates. Counter-intuitively the satellite speeds up as it descends, because v = √(GM / r) rises as r falls. A craft in a much higher orbit would coast for centuries, which is exactly why operators and regulators prefer low shells that clear themselves.
Save in your mission notebook
The labelled orbit diagram with perpendicular velocity and acceleration arrows; the full derivation of v = √(GM / r); speed and period at both shells with units; the scale table with centre-versus-surface marked; your prediction and mechanism for the engine-off question.
Parent / teacher notes
The cancellation of m in step 4 is worth dwelling on — it is the reason the feather and the hammer fall together on the Moon, and learners who see it here recognise it later. Watch for the r-versus-h error; it is the most common failure and produces an answer that is wrong by a factor of three rather than a few percent, so it is catchable. Grade 7–8 can be given GM and asked to evaluate rather than derive. Grade 9–12 should do the Kepler derivation and the projectile comparison, which needs the small-angle approximation d² / 2R.
Suggested timing: 10 minutes predict, 15 read and explain, 30 build and investigate, 25 measure and calculate, 10 independent explanation. Adapt the activities rather than rush unfinished reasoning.
Take it further
Calculate the one-way signal delay for each shell at c = 299,792 km/s: 550 ÷ 299,792 = 1.83 ms, and 35,786 ÷ 299,792 = 119 ms. A request and its reply cross the gap four times in a geostationary link, so the propagation alone costs about 477 ms before any processing. Explain why geostationary internet feels sluggish in a video call but perfectly fine for broadcast television.